Chứng minh rằng:
\(B=\sqrt{5\sqrt{5}\sqrt{5}...\sqrt{5}\sqrt{5}}+\sqrt{6+\sqrt{6}+\sqrt{6}+...\sqrt{6}+\sqrt{6}}< 8\)
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Chứng mỉnh rằng
\(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}+\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}< 8\)
\(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}=x\Rightarrow x^2=5x\Rightarrow x=5\)(n số 5)
Vậy \(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}=5\) khi \(n\rightarrow\infty\)
\(\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}=x\\ \Leftrightarrow x^2=6+\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}\\ \Leftrightarrow x^2=6+x\Rightarrow x=3\)(n số 6)
\(\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}=3\) khi \(n\rightarrow\infty\)
Vậy S < 8
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Chứng minh :
\(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5}}}}+\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}< 8\)
(2018 dấu căn )
Chứng minh: \(\sqrt{5\sqrt{5\sqrt{5\sqrt{5...\sqrt{5}}}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6...+\sqrt{6}}}}}}\le8\) có 2018 dấu căn ở mỗi hạng tử
chứng minh :a) 11+6sqrt{2} (3+sqrt{2})^2 b) sqrt{11+6sqrt{2}}+sqrt{11-6sqrt{2}}6 c) sqrt{8-2sqrt{7}}-sqrt{8+2sqrt{7}} -2 d) sqrt{49-12sqrt{5}}-sqrt{49+12sqrt{5}}-4
Đọc tiếp
chứng minh :a) 11+6\(\sqrt{2}\)= (3+\(\sqrt{2}\))\(^2\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)=6
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)= -2
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)=-4
a: \(\left(3+\sqrt{2}\right)^2=3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2\)
\(=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b: \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c: \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d: \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-2\cdot3\sqrt{5}\cdot2+4}-\sqrt{45+2\cdot3\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
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a) \(\left(3+\sqrt{2}\right)^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
c) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
d) \(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
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Chứng minh rằng các số sau đây là số nguyên:
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
B = \(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Trả lời:
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(A=\sqrt{1}\)
\(A=1\)
\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{\left(3+2\sqrt{6}+2\right).\left(49-20\sqrt{6}\right).\sqrt{3-2\sqrt{6}+2}}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{33}-11\sqrt{2}}\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right).\left(49-20\sqrt{6}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{\left(3-2\right).\left(49\sqrt{3}-60\sqrt{2}+49\sqrt{2}-40\sqrt{3}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(B=\frac{1.\left(9\sqrt{3}-11\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(B=1\)
a) Ta có: \(\sqrt{29-12\sqrt{5}}=\sqrt{20-12\sqrt{5}+9}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)
\(\Rightarrow\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{3-\left(2\sqrt{5}-3\right)}=\sqrt{3-2\sqrt{5}+3}\)
\(=\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
\(\Leftrightarrow A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)( đpcm )
CMR :\(\sqrt{5\sqrt{5\sqrt{5...\sqrt{5\sqrt{5}}}}}+\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6+\sqrt{6}}}}}< 8\)
Cho \(A=\sqrt{6+\sqrt{6...+\sqrt{6}}+\sqrt[3]{6+\sqrt[3]{6...+\sqrt[3]{6}}}}\) Chứng minh rằng 4<A<5
Chứng minh rằng:
a)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\) là số nguyên
b)\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Cho \(A=\sqrt{6+\sqrt{6...+\sqrt{6}}+\sqrt[3]{6+\sqrt[3]{6...+\sqrt[3]{6}}}}\) Chứng minh rằng 4<A<5